The proof is somehow classical. For a similar proof without the full Euler scheme expansion, one can consult [1]} Lemma 2.3 or [2]}, Theorem 10.14 and 10.26.
The proof here is an easy adaptation to the one of [3]}. Indeed, one can combine Theorem 2.2, Corollary 3.5, Remark 3.6 and Theorem 4.2 from [3]} to get the existence and uniqueness of \(\varphi \) , and points (REF ), (REF ) and (REF ). Note also that the continuity of \(\varphi \)  with respect to \(\mathbf {w}\)  is also a consequence of the previous theorems.
To construct \(\psi \)  and prove the wanted Euler expansion, let us remind the strategy of proof in [3]} (see also [6]}). Since \(\mathbf {w}\)  is in \(G^{\lfloor p \rfloor }(\mathbb {R}^d)\) , there exists an \(\mathbf {l}\in L^{\lfloor p \rfloor }(\mathbb {R}^d)\)  such that
\(\exp ^{\oplus }(\mathbf {l}) = \mathbf {w}.\) 
Remark that \(\mathbf {l} \in L^{\lfloor p \rfloor }(\mathbb {R}^d)\)  and a basis of this space is
\(e_{[i]}= e_i, \quad e_{[i,I]} = [e_i,e_{[I]}] = e_i \otimes e_{[I]} - e_{[I]}\otimes e_i.\) 
Hence
\(\mathbf {l} = \sum _{k=1}^{\lfloor p \rfloor } \sum _{I\in \lbrace 1,\cdots ,d\rbrace ^k} \mathbf {l}^{k,[I]} e_{[I]}\) 
and one can show that, in \(\mathcal {C}^{\frac{1}{p}}([0,T];C^n_{lin}(\mathbb {R}^d;\mathbb {R}^d))\) ,
\(\varphi _{s,t}=\lim _{\begin{array}{c}\pi \in \Pi ([s,t])\\ |\pi |\rightarrow 0\end{array} }\mu _{t_{N-1},t_N} \circ \cdots \circ \mu _{t_0,t_1},\) 
where \(\Pi ([s,t])\)  denotes the set of partitions of \([s,t]\) , \(\pi =\lbrace s=t_0 < t_1 \cdots < t_{N-1} < t_N=t\rbrace \)  is a partition of \((s,t)\)  and \(|\pi | = \sup _{i\in \lbrace 0,\cdots ,N-1\rbrace } |t_{i+1}-t_i|\)  is the radius of \(\pi \) , and \(\mu _{s,t} = y(1)\)  where \(y\)  is the solution of the following ordinary differential equation :
\(y^{\prime }(r) = \sum _{\begin{array}{c} k\in \lbrace 1,\cdots , \lfloor p \rfloor \rbrace  \\ I\in \lbrace 1,\cdots ,d\rbrace ^k\end{array}} \mathbf {l}^{k,[I]}_{s,t} \sigma _{[I]}(y(r)) ,\quad y(0) = x, r\in [0,1].\) 
Here, for \(i\in \lbrace 1,\cdots ,d\rbrace \) , \(\sigma _{[i] }=\sigma _i\)  and for \(k\ge 1\) , \(i\in \lbrace 1,\cdots ,d\rbrace \)  and \(I\in \lbrace 1,\cdots ,d\rbrace ^{k-1}\) , \(\sigma _{[(i,I)]} = [\sigma _i,\sigma _{[I]}] = \sigma _i \sigma _{[I]} - \sigma _{[I]} \sigma _i \) .
Now, let us define the terminal value solution of the previous equation
\(z^{\prime }(r) = \sum _{\begin{array}{c} k\in \lbrace 1,\cdots , \lfloor p \rfloor \rbrace  \\ I\in \lbrace 1,\cdots ,d\rbrace ^k\end{array}} \mathbf {l}^{k,[I]}_{s,t} \sigma _{[I]}(z(r)) ,\quad z(1) = x, r\in [0,1]\) 
and
\(\nu _{s,t}(x) = z(0).\) 
We have for all \(x\in \mathbb {R}^d\)  and all \((s,t) \in \Delta ^2_T\) ,
\(\nu _{s,t}(\mu _{s,t})(x) = \mu _{s,t}(\nu _{s,t})(x) = x,\) 
Let us define for a partition \(\pi \in \Pi ([s,t])\)
\(\nu ^{\pi }_{s,t} = \mu _{t_{0},t_1} \circ \cdots \circ \mu _{t_{N-1},t_N}\) 
and
\(\mu ^{\pi }_{s,t} = \mu _{t_{N-1},t_N} \circ \cdots \circ \mu _{t_0,t_1},\) 
then
\(\mu ^{\pi }_{s,t}(\nu ^\pi _{s,t}(x))=\nu ^{\pi }_{s,t}(\mu ^\pi _{s,t}(x))=x.\) 
Hence, one only has to prove that \(\nu ^{\pi }\)  converges to a (backward) flow in the suitable space. Note that this follows from the same arguments as the proof of Theorem 2.2 in [3]}, which guaranties that \(\psi \)  exists in \(C^n_{lin}\)  is continuous with respect to \(\mathbf {w}\)  and that for \((s,t)\in \Delta ^2_T\)  small enough (with respect to \(\mathbf {w}\) ), one has for any \(0\le m\le n\) ,
\(\sup _{x\in \mathbb {R}^d}|D^m\psi _{s,t} - D^m\nu _{s,t}(x)| \lesssim _{\mathbf {w}} (t-s)^{\frac{\lfloor p \rfloor + 1}{p}}.\) 
Furthermore, one also have (see Lemmas 3.3, 3.4 and Corollary 3.5 in [3]}), for \(a=\frac{\lfloor p \rfloor +1}{p}\)
\(\Bigg |D^m\nu _{s,t}(x)-\Bigg ((D^m id)(x)+\sum _{j=1}^{\lfloor p \rfloor }\frac{(-1)^j}{j!}\sum _{\begin{array}{c}1\le k_1,\cdots , k_j \le \lfloor p \rfloor \\ \sum _{l=1}^{j} k_l \le \lfloor p \rfloor \\ I_1 \in \lbrace 1,\cdots ,d\rbrace ^{k_1}\\ \cdots \\ I_j \in \lbrace 1,\cdots ,d\rbrace ^{k_j} \end{array}}\prod _{l=1}^j \mathbf {l}^{k_l,I_l}_{s,t}D^m\big (V_{[I_j]}\cdots V_{[I_1]}\big )(x)\Bigg )\Bigg |\lesssim _{\mathbf {w}}(t-s)^{a}. \) 
Hence, if we define \(\mathbf {v}= \exp ^{\oplus }(-\mathbf {l})\)  we have
\(\Bigg |D^m\nu _{s,t}(x) - \Bigg ((D^m id)(x) + \sum _{k=1}^{\lfloor p \rfloor } \sum _{I \in \in \lbrace 1,\cdots ,d\rbrace ^k} \mathbf {v}^{k,I}_{s,t} D^m\big (V_{I}\big )(x) \Bigg )\Bigg | \lesssim _{\mathbf {w}} (t-s)^{a}. \) 
Note first that
\(\mathbf {v}_{s,t} \otimes \mathbf {w}_{s,t} = \exp ^{\oplus }(-\mathbf {l}_{s,t}) \otimes \exp ^{\oplus }(\mathbf {l}_{s,t}) = \exp ^{\oplus }(0) = \mathbf {1} = \mathbf {w}_{s,t} \otimes \mathbf {v}_{s,t}.\) 
Furthermore, note also that for \(k> \lfloor p \rfloor \) ,
\((1-\mathbf {w}_{s,t})^{\otimes k} = 0.\) 
Hence
\(\mathbf {v}_{s,t}= \sum _{k=0}^{\lfloor p \rfloor } (\mathbf {1} - \mathbf {w}_{s,t} )^{\otimes k}\) 
and the result follows.
